# Velocity vibrations.

## Homework Statement

A mass of 0.3kg is suspended from a spring of stiffness 200Nm-1. If the mass is displaced by 10mm from its equalibrium position and released, for the resulting vibration calculate: The maximum velocity of the mass during the vibration.

v = Aωcos(wt+∅)

## The Attempt at a Solution

Now ω = 25.82cos.

so v = A 25.82cos (25.82t+∅)
So now i have to find A (amplitude) and t (displacement)
Is displacement the same as the amplitude?
Not sure on how to find the amplitude? Any guidance would be appreciated.

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Is displacement the same as the amplitude?
Not sure on how to find the amplitude?
Amplitude is nothing but the maximum displacement of the particle.
Since the mass is pulled up to 10 mm from the actual or mean position it will be the amplitude (well, while in motion the mass cannot go any further right?).
And t is not the displacement its the time. If you might recall the displacement of the mass is given by another equation. But its OK, we need the velocity here.
Remember that the velocity is maximum at the mean position, and solve....
Regards

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so the amplitude is 10mm. v = 10x25.82cos (25.82 x t x 0) because starting angle is 0? Not sure how to find time? is time 0??

rock.freak667
Homework Helper
Hint: v will be max when cos(wt+∅) is max. (So what is the maximum value of the cosine function?)

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max cos is 25.82

so 0.010x25.82cos

= 0.26cos

since vmax at t = 0

then Vmax=0.26cos(25.82x0) = 0.26ms-1?

What i dont understand is that from the velocity/amplitude graphs, velocity is at its peak when the amplitude is 0 so why have i multiplied the amplitude (displacement?) by 0.010?

rock.freak667
Homework Helper
max cos is 25.82
That isn't the max value of cos.

so 0.010x25.82cos

= 0.26cos

since vmax at t = 0

then Vmax=0.26cos(25.82x0) = 0.26ms-1?

What i dont understand is that from the velocity/amplitude graphs, velocity is at its peak when the amplitude is 0 so why have i multiplied the amplitude (displacement?) by 0.010?
Velocity is not maximum when its amplitude is zero. You were told that the amplitude is A = 0.01 m.

Velocity is maximum when the displacement is zero. This is because at the equilibrium position where x=0 all the PE of the motion has been converted into KE.

The equation is v= Aωcos(ωt+∅)
t=0 because Thats when Vmax occurs.
∅= 0 Because that is the starting angle.
So from this (ωt+∅) = 25.82

Vmax=Aωcos(ωt+∅)
so
Vmax=0.01x25.82cos(ωt+∅)
so Vmax=0.26cos(25.82t)
so max cos=0.26(25.82t)

since 25.82xt = 0

Leaves me to think the maxV= 0.26ms-1
Apologies if i am missing the obvious with this question. I am on an open university course and I seem to be struggling with this particular topic.

The equation is v= Aωcos(ωt+∅)
t=0 because Thats when Vmax occurs.
∅= 0 Because that is the starting angle.
So from this (ωt+∅) = 25.82
(ωt+∅) becomes zero not 25.82 ...
But you have taken (ωt+∅)=25.82t=0 in the coming equation, I think you substituted t=0 later, that's OK....
And the answer comes out to be right..
Just don't confuse with the equations, first retrieve all data and then substitute that's all... Regards

• 1 person